There were some grey cards and green cards. The cards were packed into 2 bags. At first, Bag S contained 145 cards and 40% of them were green cards. Bag T contained 65 cards and 80% of them were green cards. How many grey cards and green cards in total must be moved from Bag S to Bag T such that 25% of the cards in Bag S are grey and 50% of the cards in Bag T are green?

	Bag S		Bag T
Total	145		65
	Green cards	Grey cards	Green cards	Grey cards
Before	58	87	52	13
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of green cards in Bag S at first
= 40% x 145
= ⁴⁰₁₀₀ x 145
= 58

Number of grey cards in Bag S at first
= 145 - 58
= 87

Number of green cards in Bag T at first
= 80% x 65
= ⁸⁰₁₀₀ x 65
= 52

Number of grey cards in Bag T at first
= 65 - 52
= 13

Bag S in the end
25% = ²⁵₁₀₀ =¹₄
Green cards : Grey cards = 3 : 1

Bag T in the end
50% = ⁵⁰₁₀₀ =¹₂
Green cards : Grey cards = 1 : 1

Total number of green cards = 3 u + 1 p

3 u + 1 p = 58 + 52
3 u + 1 p = 110 

1 p = 110 - 3 u --- (1)

Total number of grey cards = 1 u + 1 p
1 u + 1 p = 87 + 13
1 u + 1 p = 100

1 p = 100 - 1 u --- (2)

(2) = (1)
100 - 1 u = 110 - 3 u
3 u - 1 u = 110 - 100
2 u = 10

1 u = 10 ÷ 2 = 5

Total number of grey cards and green cards that must be moved from Bag S to Bag T
= 145 - 4 u
= 145 - 4 x 5
= 145 - 20
= 125

Answer(s): 125